# How do you solve #root4(2x)+root4(x+3)=0#?

##### 1 Answer

This equation has no solutions.

#### Explanation:

Note that

So the only circumstances under which the left hand side of the given equation can be

#2x=0" "# and#" "x+3=0#

The first of these implies that

These are incompatible, so there are no real solutions.

**Complex solutions?**

Given:

#root(4)(2x) + root(4)(x+3) = 0#

Subtract

#root(4)(2x) = -root(4)(x+3)#

Raise both sides to the

#2x = x+3#

Subtract

#x = 3#

But...

#root(4)(2(color(blue)(3)))+root(4)(color(blue)(3)+3) = root(4)(6)+root(4)(6) != 0#

So there are no solutions at all.